Problem: Is ${933467}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {933467}= &&{9}\cdot100000+ \\&&{3}\cdot10000+ \\&&{3}\cdot1000+ \\&&{4}\cdot100+ \\&&{6}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {933467}= &&{9}(99999+1)+ \\&&{3}(9999+1)+ \\&&{3}(999+1)+ \\&&{4}(99+1)+ \\&&{6}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {933467}= &&\gray{9\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {9}+{3}+{3}+{4}+{6}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${933467}$ is divisible by $9$ if ${ 9}+{3}+{3}+{4}+{6}+{7}$ is divisible by $9$ Add the digits of ${933467}$ $ {9}+{3}+{3}+{4}+{6}+{7} = {32} $ If ${32}$ is divisible by $9$ , then ${933467}$ must also be divisible by $9$ ${32}$ is not divisible by $9$, therefore ${933467}$ must not be divisible by $9$.